A function y = f(x) has a second order derivative f″(x) = 6(x - 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x - 5, then the function is

Question

A function y = f(x) has a second order derivative f″(x) = 6(x - 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x - 5, then the function is (A) (x - 1)2 (B) (x - 1)3 (C) (x + 1)3 (D) (x + 1)2

Tardigrade Admin · Accepted Answer

f″ (x) = 6(x - 1) ⇒ f′ (x) = 3(x - 1)2 + c and f′ (2) = 3 ⇒ c = 0 ⇒ f (x) = (x - 1)3 + k and f (2) = 1 ⇒ k = 0 ⇒ f (x) = (x - 1)3.

Q. A function $y = f (x)$ has a second order derivative $f ″ (x) = 6 (x - 1)$ . If its graph passes through the point $(2, 1)$ and at that point the tangent to the graph is $y = 3 x - 5$ , then the function is

$(x - 1)^{2}$

$(x - 1)^{3}$

$(x + 1)^{3}$

$(x + 1)^{2}$

$f ″ (x) = 6 (x - 1) \Rightarrow f' (x) = 3 (x - 1)^{2} + c$
and $f' (2) = 3 \Rightarrow c = 0$
$\Rightarrow f (x) = (x - 1)^{3} + k$ and $f (2) = 1 \Rightarrow k = 0$
$\Rightarrow f (x) = (x - 1)^{3} .$

Q. A function y=f(x) has a second order derivative f″(x)=6(x−1). If its graph passes through the point (2,1) and at that point the tangent to the graph is y=3x−5, then the function is

(x−1)2

(x−1)3

(x+1)3

(x+1)2